The number of electrons flowing per second in the filament of a 110...

Power $(\mathrm{P})=\mathrm{V} .$\mathrm{I}$\begin{aligned} & \Rightarrow 110=(220)(\mathrm{I}) \\ & \Rightarrow \mathrm{I}=0.5 \mathrm{~A} \end{aligned}$<br />Now, $I=\frac{n \cdot e}{t}\begin{aligned} & \Rightarrow 0.5=\left(\frac{\mathrm{n}}{\mathrm{t}}\right)\left(1.6 \times 10^{-19}\right) \\ & \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{0.5}{1.6 \times 10^{-19}} \\ & \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=31.25 \times 10^{17} \end{aligned}$$$